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Chapter 9: Problem 3

The distribution of the scores on a certain exam is \(N(80,5)\) which means that the exam scores are Normally distributed with a mean of 80 and a standard deviation of 5 . a. Sketch or use technology to create the curve and label on the \(x\) -axis the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be greater than 90\. Shade the region under the Normal curve whose area corresponds to this probability.

Short Answer

Expert verified
The probability that a randomly selected score will be greater than 90 is roughly 0.0228 or 2.28%.

Step by step solution

01

Understanding Normal Distribution

A normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In this case, the mean (\(\mu\)) is 80 and the standard deviation (\(\sigma\)) is 5. These represent key points on the distribution graph.
02

Sketch the Normal Distribution Curve

Draw a curve on the x-axis. At the center of the curve is the mean (80). Then label the following points on the x-axis from left to right: \(\mu - 3\sigma,(80 - 3*5) = 65\); \(\mu - 2\sigma,(80 - 2*5) = 70\); \(\mu - \sigma,(80 - 5) = 75\); \(\mu,(80)\); \(\mu + \sigma,(80 + 5) = 85\); \(\mu + 2\sigma,(80 + 2*5) = 90\); \(\mu + 3\sigma,(80 + 3*5) = 95\)
03

Calculate the z-score

The z-score for a number \(x\) in a Normal distribution can be calculated with the following formula: \(Z = \frac{(X - \mu)}{\sigma}\). Plug in the score 90 to calculate the z-score which is \(Z = \frac{(90-80)}{5} = 2\).
04

Find the Probability That a Score Will Be Greater Than 90

We can find the area to the left of Z using Z-tables, which provides the probability that a value selected at random is less than Z. Looking up a Z of 2 in the Z-table gives us approximately 0.9772. As we need to find the probability of a score being greater than 90, that would mean finding the area to the right of Z. It can be calculated by taking 1 minus the probability of the score being less than 90 (which was found using the z-table). So, \(P(X>90) = 1- P(X<90) = 1-0.9772 = 0.0228\). This indicates that there is a 2.28% chance of a randomly selected score being greater than 90.
05

Shade the area under the curve

On your graph, shade the region to the right of the score 90. This shaded area represents the probability that a randomly selected score is greater than 90.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a fundamental concept in statistics and mathematics that deals with the likelihood or chance of an event occurring. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. When we talk about probability in a normal distribution, we refer to the area under the curve that signifies how likely the scores are to fall within a certain range.
  • In a normal distribution, such as the one with a mean of 80 and a standard deviation of 5 given in the exercise, most of the data points cluster around the mean.
  • The total area under the curve is equal to 1, representing the certainty that the score will land somewhere on the x-axis.
  • When calculating probabilities in normal distribution, we often focus on certain ranges or thresholds, like finding the likelihood of a score being greater than a particular number.
In the exercise, finding the probability that a score is greater than 90 involves determining the area under the curve to the right of 90. With the mean at 80 and using the z-score to quantify the distance of 90 from the mean, we use probability tables to find these areas.
Z-score
A Z-score is a measure used in statistics to determine the number of standard deviations a data point is from the mean of a set of data. It is a way to standardize scores on a normal distribution, making it easier to compare different data points.
  • The formula to calculate a Z-score is given by:
    \[ Z = \frac{(X - \mu)}{\sigma} \]
    where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • A Z-score tells you how far from the mean your data point is: a Z-score of 0 indicates that the data point is exactly at the mean, while positive or negative Z-scores indicate positions on either side of the mean.
In the context of the exercise, 90 was the score we were interested in. Using the formula, the Z-score calculated was 2, meaning 90 is 2 standard deviations above the mean of 80. This helps in finding the probability related to 90 by looking up Z-score tables.
Standard Deviation
Standard deviation is a measure of the spread or dispersion of a set of values. In a normal distribution, it gives us insight into how much the individual data points deviate from the mean, in essence quantifying variability.
  • In cases like the given normal distribution with a standard deviation of 5, it means that the scores typically fall within 5 units of the mean.
  • Standard deviation is a crucial parameter in normal distributions, as it plays a key role in forming the shape of the bell curve: a smaller standard deviation results in a sharper peak, while a larger value indicates a flatter spread.
  • It helps identify how much variation or "spread out" the values are from the average value (the mean).
Understanding the standard deviation allows us to predict the likelihood of various outcomes. When a score like 90 is two standard deviations away (as calculated in the previous section), it stands outside the typical range, indicating a smaller probability of occurrence in our distribution.

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Most popular questions from this chapter

Choose a test for each situation: one-sample \(t\) -test, two-sample \(t\) -test, paired \(t\) -test, and no \(t\) -test. a. A random sample of students who transfered to a 4 -year university from community colleges are asked their GPAs. Our goal is to determine whether the mean GPA for transfer students is significantly different from the population mean GPA for all students at the university. b. Students observe the number of office hours posted for a random sample of tenured and a random sample of untenured professors. c. A researcher goes to the parking lot at a large grocery chain and observes whether each person is male or female and whether they return the cart to the correct spot before leaving (yes or no).

Assume women's heights are approximately Normally distributed with a mean of 65 inches and a standard deviation of \(2.5\) inches. Which of the following questions can be answered using the Central Limit Theorem for sample means as needed? If the question can be answered, do so. If the question cannot be answered, explain why the Central Limit Theorem cannot be applied. a. Find the probability that a randomly selected woman is less than 63 inches tall. b. If five women are randomly selected, find the probability that the mean height of the sample is less than 63 inches. c. If 30 women are randomly selected, find the probability that the mean height of the sample is less than 63 inches.

The weights of four randomly and independently selected bags of tomatoes labeled 5 pounds were found to be \(5.1\), \(5.0,5.3\), and \(5.1\) pounds. Assume Normality. a. Find a \(95 \%\) confidence interval for the mean weight of all bags of tomatoes. b. Does the interval capture \(5.0\) pounds? Is there enough evidence to reject a mean weight of \(5.0\) pounds?

According to a 2018 Money magazine article, the average income in Kansas is $$\$ 53,906$$. Suppose the standard deviation is $$\$ 3000$$ and the distribution of income is rightskewed. Repeated random samples of 400 Kansas residents are taken, and the sample mean of incomes is calculated for each sample. a. The population distribution is right-skewed. Will the distribution of sample means be Normal? Why or why not? b. Find and interpret a \(z\) -score that corresponds with a sample mean of $$\$ 53,606 .$$ c. Would it be unusual to find a sample mean of $$\$ 54,500 ?$$ Why or why not?

The undergraduate grade point average (GPA) for students accepted at a random sample of 10 medical schools in the United States was taken. The mean GPA for these accepted students was \(3.75\) with a standard error of \(0.06\). The distribution of undergraduate GPAs is Normal. (Source: Accepted.com) a. Decide whether each of the following statements is worded correctly for the confidence interval. Fill in the blanks for the correctly worded one(s). Explain the error for the ones that are incorrectly worded. i. We are \(95 \%\) confident that the sample mean is between ____ \(-\) and ____. ii. We are \(95 \%\) confident that the population mean is between ____. iii. There is a \(95 \%\) probability that the population mean is between ____ and ____. b. Based on your confidence interval, would you believe that the population mean GPA is \(3.80\) ? Why or why not?
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